Thread: Basic geometry instead of measuring?

Skew,
have a look at theses two sites which provide the info below
http://www.math.harvard.edu/~ctm/gal...old/index.html
http://mathworld.wolfram.com/GoldenRatio.html

The ratio is 1: 1.618033989. or you could try the following:




Draw the square , call the midpoint of , so that . Now draw the segment , which has length
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and construct with this length. Now complete the rectangle , which is golden since
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Hi Skew!

If you know the ratio for determining a length from a given width (presumably a multiple, ie 1:1.1618 (see http://en.wikipedia.org/wiki/Golden_ratio . See also a paper on Fibonacci numbers at http://www.mcs.surrey.ac.uk/Personal.../fibInArt.html)) , then for a given length the width will be ascertained by dividing that width by the same ratio.

Edit: Ach! Ian beat me to it

Thanks fellas, but that diagram shows how I determine the length for a given width, CD in that diagram. I simply mark the width along one edge (DB) and half the width along the other (CE) then set a compass to the length of EB and scribe an arc using E as the origin. It will intersect at F, giving me the desired length.

No maths involved (apart from determining half the width ) and laid out with minimal tools in a matter of seconds.

Unfortunately, this doesn't work in reverse... given the length CF it's not possible to simply determine EB to set the compass, let alone work backwards to find CD! And yet I'm damned sure my grandfather showed me a way that it could be done. Again, with just a ruler and a compass.

I know this is one of the more "fiddly" methods, but there's a lot of similar, simpler tricks I've forgotten too. It's sad to see this knowledge disappearing, seemingly without trace.

Just as an aside, have either of you tried doing any inlay/veneer work using fibonacci or golden spirals? 'Tis far beyond my skills at the moment but I will make a coffe table that way one day. Glorious to look at and automatically the right proportions.

Gday Skew, interesting thread.

Did a google for myself & sussed out a few sites, this one may be of interest, particularly the bit under "THE NEW SOLUTION
The 1 : 2 rectangle and the Egyptian Rope Measurers "

Tried to figure it out but brain overload right now Got me intrigued, tho, will keep thinking about it


Cheers....................Sean

Thanks Scooter, the 1:2 method seems to ring a bell.

ie. Mark a "temporary width" of 1/2 the length, determine the diagonal then subtract the "temporary width" from the diagonal and use the remnant as the final width.

ARGH! My brain hurts... but I think that's it. Either that or it has to do with something else I've forgotten.

I've seen some work involving the fibonacci series (or golden spirals), but for me too that is way beyond my current skill level.

Hell's teeth, even Domi couldn't do those

I haven't drawn this out, but if I remember correctly (the famous alternative for "I think") you can sub-divide a square using the golden ratio.
So if you know CF [=(1+sq.root(5))*x] you construct a square of side CF and then subdivide the square as above, to get a BE = sq.root(5)*a*x, where a=1+sq.root(5)
You can then use proportions to find the original x

ian

Skew,
I think if you just draw in line GC on Ian's diagram, where that line intersects AB will be the width you want. Then just construct a perpendicular from AB to CD and your rectangle is right there from D to B to the other two new points. This is all provided CD and thus BD are your original lengths. There are simpler methods, but making diagrams is a pain in the butt, and I just used the one that was there.

Correction - the new rectangle is from C to A to the two newly constructed points.

Originally Posted by Skew ChiDAMN!!

Unfortunately, this doesn't work in reverse... given the length CF it's not possible to simply determine EB to set the compass, let alone work backwards to find CD! And yet I'm damned sure my grandfather showed me a way that it could be done. Again, with just a ruler and a compass.

Start with the length x,

• draw a rectangle ABCD with AB being of length x and AC of length x/2.
• Draw the diagonal AD.
• Find the point E such that DE is of length x/2.
• Find the point F such that AF is the same length as AE.
• ABFG is what you are looking for

Proportional dividers?

In another life, FIDM was experiencing communications problems. I had to take the Agilent Network Analyzer to the site and define the cause of the problem.

Due to production issues, IT would only let us insert the analyzer during lunch hour or after hours. While waiting, I was browsing in their book store. For about $4 I purchased a "Proportional Scale". (Really a circular slide rule with an added scale for proportions.)

http://www.cthruruler.com/search/search_results2.php

I have the PS 69 model and use it regularly.

UNFORTUNATELY, it is an Imperial based device.

Yeah, those proportional scales are good.

I use a pilot's navigation circular slide-rule, which works for any system.
Come to think of it, I see a lot of old-fashioned slide rules at flea markets for not much money. You could get one and make a reference mark at phi so that all of this proportional work is painless.

I know that this thread is about striking dimensions with drawing tools, and to that end I'll say that the Dover reprint of "The American Builder's Companion", published in 1827 has all of the drawing methods you'd ever want, including how to project complex mouldings. (The golden ratio is not mentioned per se, but the four orders of architecture are, and they contain this ratio. Ionic volutes are explained, and they are a derivative expression on phi)

This book was in its day the last word on proportion, ornament and style, and I get new insights every time I leaf through it.

Greg

Originally Posted by javali

Start with the length x,

• draw a rectangle ABCD with AB being of length x and AC of length x/2.
• Draw the diagonal AD.
• Find the point E such that DE is of length x/2.
• Find the point F such that AF is the same length as AE.
• ABFG is what you are looking for

Yup. Thanks Javali... it confirms what I thought, but is a more elegant way of saying it. And will look a helluva lot better when scribed into my shopnotes.

Originally Posted by Skew ChiDAMN!!

Mark a "temporary width" of 1/2 the length, determine the diagonal then subtract the "temporary width" from the diagonal and use the remnant as the final width.

Sometimes it's just not easy putting diagrams into words.

Originally Posted by gregoryq

I know that this thread is about striking dimensions with drawing tools, and to that end I'll say that the Dover reprint of "The American Builder's Companion", published in 1827 has all of the drawing methods you'd ever want, including how to project complex mouldings. (The golden ratio is not mentioned per se, but the four orders of architecture are, and they contain this ratio. Ionic volutes are explained, and they are a derivative expression on phi)

This book was in its day the last word on proportion, ornament and style, and I get new insights every time I leaf through it.

Hmmm... interesting. (One google search later) I assume this is the one by Asher Benjamin?